Today I tried to teach a high school student about inverse functions. I gave him this problem and defined the parts that he didn't understand:


$f: \mathbb{R} \to \mathbb{R}$, $x \mapsto x^2 + 1$

$g: [0, \infty) \to [1, \infty)$, $x \mapsto x^2 + 1$

$h: [0, \infty) \to (-\infty, \infty)$, $x \mapsto x^2 + 1$

$p: \{..., -5/2, -3/2, -1/2, 0, 1, 2, ... \} \to \mathbb{R}$, $x \mapsto x^2 + 1$

$q: \mathbb{R} \to \mathbb{R}$, $x \mapsto 3x + 2$

(a) Which of these functions are one-to-one?

(b) Which of these functions are onto?

(c) Which of these functions have inverses?

(d) What are the inverses? Over what domains are the inverses defined?

(e) For the functions that have inverses, evaluate those inverses at $x = -1$.

However, I think he found the problem too difficult. What would be better problems to give him? More generally, what could I have done to help him better understand inverse functions?


  • 1
    $\begingroup$ Three of the four functions that have inverses can’t be evaluated at $-1$, because $-1$ isn’t in the domain of the inverse. The symbol $x$ is not tied to the domain of the original function. $\endgroup$ Dec 24, 2011 at 21:07

2 Answers 2


When introducing the ideas of one-to-one and onto to my students, I like to begin with some simple pictures so they can "see" what's going on. For example,

onto and not

Draw a nice little picture of an onto function (the one I've drawn isn't one-to-one). Then add some more dots in the codomain and voila! It's not onto anymore. Draw a few pictures to illustrate one-to-one as well. To picture inverse functions draw a one-to-one, onto function and then "flip" all of the arrows and you've got the inverse.

Draw a bunch of pictures adding and deleting points to make functions one-to-one and/or onto or not.

Next, I'd play around with the example: $f(x)=x^2$.

1) $f: \mathbb{R} \to \mathbb{R}$ is neither one-to-one or onto.

Why? Showing something fails to be onto is simple, merely find something in the codomain that doesn't get "hit". So for example $-1$ is not in the range. When we try to solve the equation $f(x)=x^2=-1$, the only solutions we get are $x = \pm \sqrt{-1}$ (which fail to be in our domain). So nothing maps to $-1$. Thus it's not in the range so not onto.

2) Then ask, "How could we 'fix' the function so it is onto?"

$f: \mathbb{R} \to \mathbb{R}_{\geq 0} = [0,\infty)$. Now the equation $f(x)=x^2=y$ can always be solved: $x = \pm \sqrt{y}$ (which are real since $y \geq 0$.

Next, note that there are two solutions for each (non-zero) $y$. So the function isn't one-to-one.

3) How can we "fix" $f$ again so that it's one-to-one?

$f: [0,\infty) \to [0,\infty)$

Now $f(x)=f(y)$ so $x^2=y^2$ implies that $x=y$.

4) So finally, $f: [0,\infty) \to [0,\infty)$ defined by $f(x)=x^2$ is bijective (one-to-one and onto). How can we find an inverse? Go back to what N.S. suggests: it's all about solving for "x" given the equation $y=x^2$. We solve $x=\pm \sqrt{y}$ remembering that we're only allowed non-negative solutions $x=\sqrt{y}$. So $f^{-1}:[0,\infty) \to [0,\infty)$ is defined by $f^{-1}(x)=\sqrt{x}$.

Make sure to check $f(f^{-1}(x))=(\sqrt{x})^2=x$ (the inverse works on the right because $f$ is onto) and $f^{-1}(f(x))=\sqrt{x^2}=x$ (the inverse works on the left because $f$ is one-to-one).

I hope this gives you a few ideas to help get started. :)

  • 2
    $\begingroup$ (+1) Regarding pictures and flipping, another good one is to take a one-to-one function and draw the inverse by reflecting the graph across the line $y=x$. Doing this with non one-to-one functions then also proves instructive, and even provides an opportunity to reinforce the definition (and intuitive notions) of a function. $\endgroup$
    – cardinal
    Dec 24, 2011 at 20:55
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    $\begingroup$ Good idea. I neglected the geometric side of things because I typically think of teaching this stuff at the beginning of an abstract algebra course. When working with functions defined on subsets of the reals, certainly seeing properties like one-to-one (passing the horizontal line test) can be quite helpful. $\endgroup$
    – Bill Cook
    Dec 24, 2011 at 21:34
  • $\begingroup$ Thank you for the helpful and detailed answer! $\endgroup$
    – badatmath
    Dec 26, 2011 at 3:00

The approach I find most intuitive is via equations.

Ask him to solve the equation $x^2+1=y$ for $x$, and then ask him if the answer he gets is a function...

Start building for there:

For which $y$ does this equation have solution(s)? When is the solution a function (i.e., unique)? Write the solution as a function in $y$... Which is the largest domain? What is the range?

What is the connection between this function, domain , range and the original function?

  • 4
    $\begingroup$ The problem with this approach is that it can easily encourage students to think of finding the inverse as a mechanical operation, just another mysterious manipulation of symbols followed by an even more mysterious interchange of $x$ and $y$. $\endgroup$ Dec 24, 2011 at 21:04
  • $\begingroup$ Thanks! Yeah that was the way I learned it in middle school... $\endgroup$
    – badatmath
    Dec 26, 2011 at 3:00

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